# Compensation Theorem : Working, Examples & Its Applications

In network theory, it is very significant to study or know the effect of change within impedance in one of its branches. So it will affect the corresponding currents & voltage of the circuit or network. So the compensation theorem is used to know the change within the network. This network theorem simply works on Ohm’s law concept which states that, whenever current is supplied throughout the resistor, then some amount of voltage will drop across the resistor. So this voltage drop will resist the voltage source. Thus, we connect an additional voltage source in reverse polarity contrasted to the voltage source & the magnitude is equivalent to the voltage drop. This article discusses an overview of a compensation theorem – working with applications.

## What is Compensation Theorem?

The compensation theorem in network analysis can be defined as; in a network, any resistance can be replaced with a voltage source that includes zero internal resistance & a voltage equivalent to the voltage drop across the replaced resistance because of the flowing current throughout it.

Let’s assume the flow of current ‘I’ throughout that ‘R’ resistor & voltage drops because of this flow of current across the resistor is (V = I.R). Based on the compensation theorem, this resistor is replaced through a voltage source that generates voltage & which will be directed against the network voltage direction or current direction.

### Compensation Theorem Solved Problems

The example problems of the compensation theorem are given below.

#### Example1:

For the following circuit

1). Find the current flow throughout the AB branch once the resistance is 4Ω.
2). Find the flow of current throughout the AB branch with compensation theorem once the resistance 3Ω is changed with 9Ω.
3). Verify the compensation theorem.

Solution:

As shown in the above circuit, the two resistors like 3Ω & 6Ω connected in parallel, and also this parallel combination is simply connected with the 3Ω resister in series then, equal resistance will be;

Re1 = 6 || 3 + 3 => (6×3/6+3) + 3
= (18/9) + 3 => 2+3 = 5 Ω.

Based on Ohm’s law;

8 = I (5)
I = 8 ÷ 5
I = 1.6 A

Now, we have to find the flow of current throughout the AB branch. Thus, based on the rule of the current divider;

I’ = 1.6 (6)/6+3 => 9.6/9 = 1.06A

2). Now we have to change the 3Ω resistor with a 9Ω resistor. Based on the compensation theorem, we should include a new voltage source within series with the 9Ω resistor & the voltage source value is;

VC = I’ ΔZ

Where,

ΔZ = 9 – 3 = 6 Ω & I’ = 1.06 A.

VC = (1.06) x 6 Ω = 6.36V

VC = 6.36V

The modified circuit diagram is shown below.

Now we have to find the equivalent resistance. So, the resistors like 3Ω &6Ω are simply connected in parallel. After that this parallel combination is simply connected in series by a 9Ω resistor.

Req = 3||6+9

Req = (3×6||3+6) +9

Req = (18||9) +9

Req = (2) +9

Req = 11ohms

Based on Ohm’s law;

V = ΔI x R

6.36 = ΔI (11)

ΔI = 6.36 ÷ 11

ΔI = 0.578 A

Thus, based on the compensation theorem; the change within the current is 0.578 A.

3). Now we have to prove the compensation theorem by calculating the flow of current in the following circuit with a 9Ω resister. So, the modified circuit is given below. Here, resistors like 9Ω & 6Ω are connected in parallel and this combination is simply connected in series by the 3Ω resistor.

REq = 9 | | 6 + 3

REq = (6×9 | 6 + 9) + 3

REq = (54 | 15) + 3

REq = 45+54/15 => 99/15 => 6.66ohms

From the circuit above

8 = I (6.66)

I = 8 ÷ 6.66

I = 1.20A

Based on the current divider rule;

I’’ = 1.20 (6)/6+9

I’’ = 1.20 (6)/6+9 =>7.2/15 =>0.48A

ΔI = I’ – I”

ΔI = 1.06-0.48 = 0.578A

Therefore, the compensation theorem is proved that the change within current is calculated from the theorem which is similar to the change within current measured from the actual circuit.

#### Example2:

The resistance value in the two terminals of the following circuit A & B is modified to 5ohms then what is the compensation voltage?

For the above circuit, first, we need to apply KVL

-8+1i+3i =0

4i = 8 => I = 8/4

I = 2A

ΔR = 5Ω – 3Ω

ΔR = 2Ω

The compensation voltage is

Vc = I [ΔR]

Vc = 2×2

Vc = 4V

### Compensation Theorem in AC Circuits

Find the current flow change within the following AC circuit if a 3 ohms resistor is replaced through a 7ohms resistor with the compensation theorem and also prove this theorem.

The above circuit includes only resistors as well as separate current sources. Thus, we can apply this theorem to the above circuit. So this circuit is supplied through a current source. So now we have to find the flow of current throughout the branch of the 3Ω resistor with the help of KVL or KCL. Although, this flow of current can found easily by using the current divider rule.

So, based on the current divider rule;

I = (8(7)/7+3) A => 56/10A => 5.6A.

In the actual circuit with a 3ohms resistor, the flow of current throughout that branch is 7A. So we have to change this 3ohm resistor with 7ohm. Because of this change, the flow of current throughout that branch will also be changed. So now we can find this current change with the compensation theorem.

For that, we have to design a compensation network by removing all the available independent sources within the network by simply open-circuiting the current source & short-circuiting the voltage source. In this circuit, we have only a single current source which is an ideal current source. So, we do not need to include the inside resistance. For this circuit, the next modification we need to do is to include an additional voltage source. So this voltage value is;

VC = I ΔZ => 7 × (7 – 3)

VC = 7 × 4 => 28 V

Now the compensation circuit with a voltage source is shown below.

This circuit includes only a single loop where the current supplies throughout the 7Ω branch will provide us the flow of current change i.e,(∆I).

ΔI = VC ÷ (7+7) => 28 ÷ 14 => 2 A

For proving this theorem, we have to find the flow of current within the circuit by connecting a 7Ω resistor as shown in the circuit below.

I” = (8 (7)) ÷ (7 + 7)

I” = 56 ÷ 14

I” = 4 A

Now apply the current divider rule;

To find the change in current, we need to subtract this current from the current that passes through the original network.

ΔI = I – I”

ΔI = 7 – 4 => 3 A

Therefore, the compensation theorem is proved.

#### Why do We Need a Compensation Theorem?

• The compensation theorem is very useful because it provides information regarding the change within the network. This network theorem also allows us to find out the exact current values within any branch of a network once the network is substituted directly to any specific change in a single step.
• By using this theorem we can get the approximate effect of minute changes within the elements of a network.

The advantages of the compensation theorem include the following.

• The compensation theorem provides information regarding the change within the network.
• This theorem works on Ohm’s law basic concept.
• It helps in discovering the changes within voltage or current once the resistance value is adjusted within the circuit.

### Applications

The applications of the compensation theorem include the following.

• This theorem is frequently utilized in obtaining the approximate small changes effect within the electrical network elements.
• This is very useful particularly for analyzing the bridge network’s sensitivity.
• This theorem is used to analyze the networks where the branch elements’ values are changed & also for studying the tolerance effect on such values.
• This allows you to determine the right current values within any networked branch once the network is directly substituted to any specific change within a single step.
• This theorem is the most significant theorem within network analysis which is used for calculating the sensitivity of the electrical network and solving electrical networks & bridges.

Thus, this is an overview of a compensation theorem in network analysis – example problems and their applications. So in this network theorem, the resistance in any circuit can be changed by a voltage source, having a similar voltage when the voltage drops across the resistance which is changed. Here is a question for you, what is the superposition theorem?